Models of the Hydrogen Atom

Start on Bohr and step n from 1 to 6. That is the whole model: circular orbits at rₙ = n²a₀, one per allowed energy, and an electron that is somewhere definite at every moment. It got the hydrogen spectrum right in 1913, which is why anyone remembers it.

Now switch to Schrödinger at the same n. The energy readout does not move. The orbit does — it is gone, replaced by a cloud with no edge and, for l > 0, no spherical symmetry at all. Then step l and m and watch shapes appear that the Bohr model has no vocabulary for.

n sets the size and the energy: ⟨r⟩ = ½(3n² − l(l+1)) a₀, so 4s is a long way outside 2s. l sets the shape — 0 is a sphere, 1 is a dumbbell, 2 is the four-lobed clover. m only turns it: p_x, p_y and p_z are one object pointing three different ways, with exactly the same energy — until a magnetic field picks out an axis and splits them apart, which is where “magnetic quantum number” got its name.

One honesty note about the shapes. The states with a definite m are complex, and |ψ|² for them has no dependence on the azimuthal angle at all — every m ≠ 0 orbital would render as the same featureless doughnut. What is drawn here are the realcombinations, the ones chemistry uses. They span the same space, have the same energy and the same node count, and they are the version with shapes.

The nodes are the give-away, and they are countable. Every orbital has exactly n − 1 nodal surfaces. l of them are angular (flat planes or cones through the nucleus, where the colour flips), and the remaining n − l − 1 are radial (whole spherical shells of zero probability). Set n=3, l=0 and turn on the cutaway: three nested shells of density, separated by two spherical surfaces the electron is never found on — and no route from the innermost shell to the outermost that avoids crossing them. A particle on an orbit cannot do that. A standing wave can.

Point a hand spectroscope at a hydrogen lamp and you see four sharp lines on a black field: red, cyan, blue-violet, violet. Those are the Balmer series — every jump that ends on n = 2 — and nothing else about hydrogen is visible to the naked eye. Every Lyman line (ending on n = 1) is ultraviolet; every Paschen line is infrared.

These are vacuum wavelengths, computed from the Rydberg formula with the reduced-mass Rydberg constant for hydrogen. A lab handbook usually quotes them in air, which shortens each by about 0.03% — the familiar 656.28 nm for the red line is the same photon as the 656.46 nm above.

You changed the picture of the atom completely and the spectrum did not move. Both models put the energy at −13.6 eV / n², so both predict the Balmer lines to four figures, and for thirteen years that was enough to make Bohr’s orbits look correct.

What broke them was everything the spectrum does not depend on. Bohr had no l, so he could not say why lines split in a magnetic field, why some transitions are bright and others forbidden, or why the periodic table has the shape it has. Schrödinger’s version has three quantum numbers instead of one, and the two extra ones are exactly the ones that turned out to explain chemistry — while contributing nothing at all to the energy of a hydrogen atom.

One more thing worth sitting with: nothing on this page is a photograph. The cloud is a probability density, so a single measurement returns a single position, and it is only the ensemble of many measurements that has the shape you rotated. The Bohr orbit is the honest, wrong version of the same information — a definite path that happens to reproduce the right energies.